A triangle has sides A, B, and C. The angle between sides A and B is #pi/4# and the angle between sides B and C is #pi/12#. If side B has a length of 5, what is the area of the triangle?

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Answer 1 · 2017-06-14T08:02:41

The area is #=2.64u^2#

The angle between #A# and #B# is

#theta=pi-(1/12pi+1/4pi)=pi-4/12pi=2/3pi#

#sintheta=sin(2/3pi)=sqrt3/2#

We apply the sine rule to find #=A#

#A/sin(1/12pi)=B/sin(2/3pi)#

#A=Bsin(1/12pi)/sin(2/3pi)#

The area of the triangle is

#a=1/2ABsin(1/4pi)#

#a=1/2Bsin(1/12pi)/sin(2/3pi)*B*sin(1/4pi)#

#=1/2*5*5*sin(1/12pi)*sin(pi/4)/sin(2/3pi)#

#=2.64#

Answer 2 · 2017-06-14T08:22:41

#2.64# #unit^2#

Let say the angle, #a = pi/12 and c = pi/4#. Then angle between C and A, #b# #= pi - 1/4 pi - 1/12 pi = 8/12 pi = 2/3 pi#.

Area of triangle, # = 1/2 BC sin a#
#= 1/2 (5)(C) sin (pi/12)#.#->i#

use, #B/sin b = C/sin c# to find C

#5/(sin (2/3 pi)) = C /(sin (1/4 pi))#

#C =5/(sin (2/3 pi)) * sin (1/4 pi) = 4.08#unit

Area of triangle, plug in #C# in #->i#
#= 1/2 (5)(4.08) sin (pi/12) = 2.64# #unit^2#